2015多校第一场1008 hdu 5295 Unstable

题意：给定AB，BC,CD,DA距离，已经AB的中点E，CD的中点F的距离EF。给出一个合法的A,B,C,D坐标。

做法：和题解一样。首先其中B，C两个点的坐标可以自己定，不妨设C为（0,0），B为（BC，0），然后从A点过F作A'令AF=A'F，由于对顶三角形的原因，A'C=AD。由于相似三角形的原因A'B=2*EF，已知两点坐标和三边长，可以求出A'的坐标。然后作点G，令向量GB=向量A'C,这样就可以连接GC,GD,GC已知。由于BG与AD此时是两条平行且相等的边，所以四边形ADGB构成了一个平行四边形，所以GD=AB。所以可以通过GC，GD，CD已经点G，C的坐标求出点D坐标。那么A的坐标就很容易求了。

PS：上面两个求三角形的过程中，要考虑不是三角形的情况。

PSS：标程的求圆交点姿势比我不知道高到哪里去了。于是就学习了。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
using namespace std;
#define eps 1e-6
int sgn(double x){
	if(fabs(x) <= eps) return 0;
	return x > 0 ? 1 : -1;
}
struct Point{
	double x,y;
	Point(){}
	Point(const double  &x,const double  &y):x(x),y(y){}
	double sqrlen()
	{
		return x*x+y*y;
	}
	double len()
	{
		return sqrt(sqrlen());
	}
	void print()
	{
		printf("%.6lf %.6lf\n",x,y);
	}
};
Point operator +(Point a,Point b)
{
	return Point(a.x+b.x,a.y+b.y);
}
Point operator -(Point a,Point b)
{
	return Point(a.x-b.x,a.y-b.y);
}
struct Circle{
	Point o;
	double r;
	Circle(){}
	Circle(const Point &o,const double &r):o(o),r(r){}
};
bool zero(double x)
{
	if(fabs(x)<=eps)return true;
	return false;
}
bool operator == (Circle a,Circle b)
{
	Point c=a.o-b.o;
	if(zero(c.x)&&zero(c.y)&&zero(a.r-b.r))return true;
	return false;
}


void c2point(Circle c1,Circle c2,Point &rp1,Point &rp2){
	if(c1 == c2){
		rp1 = c1.o + Point(0,c1.r);
		rp2 = c1.o - Point(0,c1.r);
		return;
	}
	Point p1 = c1.o , p2 = c2.o;
	double r1 = c1.r , r2 = c2.r;
	double a = p2.x - p1.x , b = p2.y - p1.y , r = (a*a + b*b + r1*r1 - r2*r2) / 2;
	double tmp;
	if(a == 0 && b != 0){
		rp1.y = rp2.y = r / b;
		tmp = r1 * r1 - rp1.y * rp1.y;
		if(sgn(tmp)<=0) tmp = 0;
		rp1.x = sqrt(tmp);
		rp2.x = -rp1.x;
	}
	else if(a != 0 && b == 0){
		rp1.x = rp2.x = r / a;
		tmp = r1 * r1 - rp1.x * rp1.x;
		if(sgn(tmp)<=0) tmp = 0;
		rp1.y = sqrt(tmp);
		rp2.y = -rp1.y;
	}
	else if(a != 0 && b != 0){
		double delta = b*b*r*r - (a*a + b*b) * (r*r - r1*r1*a*a);
		if(sgn(delta)<=0) delta = 0;
		rp1.y = (b*r + sqrt(delta)) / (a*a + b*b);
		rp2.y = (b*r - sqrt(delta)) / (a*a + b*b);
		rp1.x = (r - b*rp1.y) / a;
		rp2.x = (r - b*rp2.y) / a;
	}
	rp1.x += p1.x , rp1.y += p1.y;
	rp1.x += p1.x , rp2.y += p1.y;
}
Point A,B,C,D,_A;
double AB , BC , CD , DA , EF;
int main()
{
//	freopen("1008.in","r",stdin);
//	freopen("myout.out","w",stdout);
	int T,cas=0;
	scanf("%d",&T);
	while(T--)
	{
		cas++;
		scanf("%lf%lf%lf%lf%lf",&AB,&BC,&CD,&DA,&EF);
		B=Point(BC,0);
		C=Point(0,0);
		Point nouse;
		Circle BO,CO;
		BO=Circle(B,2*EF);
		CO=Circle(C,DA);
		c2point(CO,BO,_A,nouse);
		Point G=_A+(B-C);
		Circle GO;
		CO=Circle(C,CD);
		GO=Circle(G,AB);
		c2point(CO,GO,D,nouse);
		A=D+(C-_A);
		printf("Case #%d:\n",cas);
		A.print();
		B.print();
		C.print();
		D.print();
	}
}